The equations of the common tangents to the e | vedclass

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Question

Equation of tangent to above curves are respectively.

$y ^{2}= mx +\frac{1}{ m }$ and $y = mx +\sqrt{8 m ^{2}+2}$

Comparing $\frac{1}{ m }=\sqrt

Vedclass · Accepted Answer

both $(A)$ and $(B)$

Vedclass · Answer

Equation of tangent to above curves are respectively.

$y ^{2}= mx +\frac{1}{ m }$ and $y = mx +\sqrt{8 m ^{2}+2}$

Comparing $\frac{1}{ m }=\sqrt{8 m ^{2}+2}$

$\Rightarrow m ^{2}\left(8 m ^{2}+2\right)=1$

seeing the options

$m =\pm \frac{1}{2}$ satisfy the equation

$\Rightarrow y=\pm \frac{1}{2} x \pm 2 \Rightarrow 2 y=\pm x \pm 4$

i.e. $2 y=x+4 \& x+2 y+4=0$

The equations of the common tangents to the ellipse, $ x^2 + 4y^2 = 8 $ $\&$ the parabola $y^2 = 4x$ can be

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